question_answer

If ABCDE is a regular pentagon, then the angle BDE is equal to

A) \[90{}^\circ \]

B) \[72{}^\circ \]  

C) \[60{}^\circ \]

D)        \[54{}^\circ \]

Correct Answer: B

Solution :

 Since side of pentagon, \[x=5\] \[\therefore \] Each interior \[=\frac{2\times 5-4}{5}\times {{90}^{o}}={{108}^{o}}\] Since \[\Delta \,\,CDE\] is an isosceles triangle \[\therefore \]  \[2\angle CDB=180-108=72\] or            \[\angle CDE={{36}^{o}}\] \[\therefore \]  \[\angle BDE=108-{{36}^{o}}={{72}^{o}}\]